At certain loading conditions, back e.m.f. in DC motor was found half of the supply voltage. Then power delivered by DC motor is

Relation between Mechanical power (P_m), Supply Voltage (V_t), and Back emf (E_b):

The back emf in the dc motor is expressed as:

E_b = V – I_aR_a .......(1)

E_b = back emf

I_a = armature current

V_t = terminal voltage

R_a = resistance of amature

The Power developed on the motor is expressed by

P_m = E_bI_a = VI_a – I_a²Ra ......(2)

On differentiating of the given equation

\(\frac{{{\rm{d}}{{\rm{P}}_{\rm{m}}}}}{{{\rm{d}}{{\rm{I}}_{\rm{a}}}}} = \frac{{\rm{d}}}{{{\rm{d}}{{\rm{I}}_{\rm{a}}}}}\left( {{\rm{V}}{{\rm{I}}_{\rm{a}}} - {\rm{I}}_{\rm{a}}^2{{\rm{R}}_{\rm{a}}}} \right)\)

For maximum power develop

\(\frac{{{\rm{d}}{{\rm{P}}_{\rm{m}}}}}{{{\rm{d}}{{\rm{I}}_{\rm{a}}}}} = {\rm{V}} - 2{{\rm{I}}_{\rm{a}}}{{\rm{R}}_{\rm{a}}} = 0\)

V = 2 I_aR_a ⇒ I_aR_a = V / 2

From the back emf equation 

E_b = V – I_aR_a = V – V / 2

\({{\rm{E}}_{\rm{b}}} = \frac{{\rm{V}}}{2}\) .......(3)

The maximum power is developed in the motor when the back emf is equal to half of the supply voltage.

and the maximum power is

(P_m)_max = VI_a – I­_a²R_a = I_a (V – I­_aR_a)

\({\left( {{{\rm{P}}_{\rm{m}}}} \right)_{{\rm{max}}}} = \frac{{{\rm{V}}{{\rm{I}}_{\rm{a}}}}}{2}\)

(P_m)_max = E_bI_a

Key Points

Back EMF in DC Motor:

• When the current-carrying conductor placed in a magnetic field, the torque induces on the conductor, the torque rotates the conductor which cuts the flux of the magnetic field.
• According to the Electromagnetic Induction Phenomenon “when the conductor cuts the magnetic field, EMF induces in the conductor”.
• It is seen that the direction(Right hand rule) of the induced emf is opposite to the applied voltage. Thereby the emf is known as the counter emf or back emf.